The radius of the Sun is 6.95 × 10^8 m. It radiates heat atthe rate of 5.32 x 10

Question

The radius of the Sun is 6.95 × 10^8 m. It radiates heat atthe rate of 5.32 x 10^26 W. Assuming that it is a perfect emitter,what is the temperature of the surface of the Sun? TheStefan-Boltzmann constant is 5.67 x 10^-8W/(m2·K4).   A) 6.27 × 10^3K
B) 8.25 × 103 K
C) 8.87 × 10^3 K D) 3.93 × 10^7K
E) 5.78 × 10^7 K Please give step by ste solition with answer A) 6.27 × 10^3K
B) 8.25 × 103 K
C) 8.87 × 10^3 K D) 3.93 × 10^7K
E) 5.78 × 10^7 K Please give step by ste solition with answer Please give step by ste solition with answer

Explanation / Answer

Given Radius of Sun.,R = 6.95 × 10^8 m => Area of Sun, A = (4)R2 = 606.68*1016 m2 Power, P = 5.32 x 10^26 W Temperature, T = ? Stefan-Boltzmann constant, = 5.67 x 10^-8W/(m2·K4) As the body is Perfect Emmiter => emissivity ,e = 1 Now ,According to Stefan-Boltzman law, P = AeT4 => 5.32 x 1026 = 606.68* 1016 *1*5.67 x10-8 *T4 => T4 =1.55* 10-3 * 1026*10-16 *108 = 1.55 *1015 => T = 6.274 * 10^3 K

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